已经连咕了好几天博客了；比较经典的题目
题目大意
给出一个 N 个点的树和$K_i$， 求每个点到其他所有点距离中第 $K_i$ 小的数值。
题目分析
做法一：点分树上$log^3$
首先暴力做法：对于每个节点维护其他点距离的值域线段树。这个做法的瓶颈在于关于边$(u,v)$线段树的转移。那么可以利用点分树（为了保证复杂度）换一种容斥的思路利用重复的信息：记$f_i$为以$i$为根的点分树内所有其他点到点$i$的距离的值域线段树；$g_i$为以$i$为根的点分树内，所有点到$i$的点分树父亲的距离 的值域线段树。
询问的时候，记$LIM$为二分的长度，查询点为$pos$，$lst$的点分树父亲为$i$，那么每一层的贡献就是$f_i$中$LIM-dis(pos,i)$减去$g_{lst}$中$LIM-dis(pos,i)$的代价，注意还要特判一下$(pos,i)$这条路径是否会被算入贡献。
1 #include<bits/stdc++.h> 2 const int maxLog = 20; 3 const int maxn = 100035; 4 const int maxm = 200035; 5 const int maxNode = 20000035; 6 7 struct node 8 { 9 int l,r,val; 10 }; 11 int LIM; 12 struct RangeSeg 13 { 14 int tot,rt[maxn]; 15 node a[maxNode]; 16 void modify(int &rt, int l, int r, int pos) 17 { 18 if (!rt) rt = ++tot; 19 ++a[rt].val; 20 if (l==r) return; 21 int mid = (l+r)>>1; 22 if (pos <= mid) modify(a[rt].l, l, mid, pos); 23 else modify(a[rt].r, mid+1, r, pos); 24 } 25 int query(int rt, int L, int R, int l, int r) 26 { 27 if (!rt) return 0; 28 if (L <= l&&r <= R) return a[rt].val; 29 int mid = (l+r)>>1, ret = 0; 30 if (L <= mid) ret = query(a[rt].l, L, R, l, mid); 31 if (R > mid) ret += query(a[rt].r, L, R, mid+1, r); 32 return ret; 33 } 34 void modify(int x, int c) 35 { 36 modify(rt[x], 1, LIM, c); 37 } 38 int query(int x, int l, int r) 39 { 40 if (l <= r) return query(rt[x], l, r, 1, LIM); 41 else return 0; 42 } 43 }f,g; 44 int n,k[maxn]; 45 int dep[maxn],fat[maxn][maxLog],lay[maxn]; 46 int size[maxn],bloTot,son[maxn],root; 47 int edgeTot,head[maxn],nxt[maxm],edges[maxm]; 48 bool vis[maxn]; 49 50 int read() 51 { 52 char ch = getchar(); 53 int num = 0, fl = 1; 54 for (; !isdigit(ch); ch=getchar()) 55 if (ch=='-') fl = -1; 56 for (; isdigit(ch); ch=getchar()) 57 num = (num<<1)+(num<<3)+ch-48; 58 return num*fl; 59 } 60 void addedge(int u, int v) 61 { 62 edges[++edgeTot] = v, nxt[edgeTot] = head[u], head[u] = edgeTot; 63 edges[++edgeTot] = u, nxt[edgeTot] = head[v], head[v] = edgeTot; 64 } 65 int lca(int x, int y) 66 { 67 if (dep[x] > dep[y]) std::swap(x, y); 68 for (int i=17; i>=0; i--) 69 if (dep[fat[y][i]] >= dep[x]) y = fat[y][i]; 70 if (x==y) return x; 71 for (int i=17; i>=0; i--) 72 if (fat[x][i]!=fat[y][i]) x = fat[x][i], y = fat[y][i]; 73 return fat[x][0]; 74 } 75 int dist(int x, int y){return dep[x]+dep[y]-(dep[lca(x, y)]<<1);} 76 void dfs(int x, int fa) 77 { 78 fat[x][0] = fa, dep[x] = dep[fa]+1; 79 for (int i=1; i<=17; i++) 80 fat[x][i] = fat[fat[x][i-1]][i-1]; 81 for (int i=head[x]; i!=-1; i=nxt[i]) 82 if (edges[i]!=fa) dfs(edges[i], x); 83 } 84 void getRoot(int x, int fa) 85 { 86 size[x] = 1, son[x] = 0; 87 for (int i=head[x]; i!=-1; i=nxt[i]) 88 { 89 int v = edges[i]; 90 if (v==fa||vis[v]) continue; 91 getRoot(v, x); 92 size[x] += size[v]; 93 son[x] = std::max(son[x], size[v]); 94 } 95 son[x] = std::max(son[x], bloTot-size[x]); 96 if (son[x] < son[root]) root = x; 97 } 98 void color(int Top, int x, int fa) 99 { 100 if (Top!=x) f.modify(Top, dist(Top, x)); 101 if (lay[Top]) g.modify(Top, dist(lay[Top], x)); 102 for (int i=head[x]; i!=-1; i=nxt[i]) 103 if (edges[i]!=fa&&!vis[edges[i]]) color(Top, edges[i], x); 104 } 105 void deal(int x, int pre) 106 { 107 lay[x] = pre, color(x, x, x), vis[x] = true; 108 for (int i=head[x]; i!=-1; i=nxt[i]) 109 { 110 int v = edges[i]; 111 if (vis[v]) continue; 112 bloTot = size[v], root = 0, getRoot(v, 0); 113 deal(root, x); 114 } 115 } 116 int count(int x, int num) 117 { 118 int ret = f.query(x, 1, num); 119 for (int i=lay[x],lst=x; i; lst=i,i=lay[i]) 120 { 121 int d = dist(x, i); 122 if (d <= num) ++ret, ret += f.query(i, 1, num-d)-g.query(lst, 1, num-d); 123 } 124 return ret; 125 } 126 int main() 127 { 128 memset(head, -1, sizeof head); 129 LIM = n = read(); 130 for (int i=1; i<=n; i++) k[i] = n-read(); 131 for (int i=1; i<n; i++) addedge(read(), read()); 132 dfs(1, 0); 133 bloTot = n, root = 0, son[0] = n, getRoot(1, 0); 134 deal(root, 0); 135 for (int i=1; i<=n; i++) 136 { 137 int L = 0, R = LIM, ans = 0; 138 for (int mid=(L+R)>>1; L<=R; mid=(L+R)>>1) 139 if (count(i, mid) < k[i]) L = mid+1, ans = mid; 140 else R = mid-1; 141 printf("%d ",ans); 142 } 143 return 0; 144 }
做法二：序列问题分块
不说了。类似的套路见#6046. 「雅礼集训 2017 Day8」爷